If
Question:

If $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$, then $x=$

(a) $\frac{1}{2}$

(b) $\frac{\sqrt{3}}{2}$

(c) $-\frac{1}{2}$

(d) none of these

Solution:

(b) $\frac{\sqrt{3}}{2}$

We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$.

$\therefore \sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$

$\Rightarrow \frac{\pi}{2}-\cos ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$

$\Rightarrow-2 \cos ^{-1} x=\frac{\pi}{6}-\frac{\pi}{2}$

$\Rightarrow-2 \cos ^{-1} x=-\frac{\pi}{3}$

$\Rightarrow \cos ^{-1} x=\frac{\pi}{6}$

$\Rightarrow x=\cos \frac{\pi}{6}$

$\Rightarrow x=\frac{\sqrt{3}}{2}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.